NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry

How to present this for full marks. Write the formula once, then the substitution, then the arithmetic, then the simplified surd.

• Part (i): the two points differ by 2 across and 2 down, so the gap is √2²+2²=√8. Always pull out the largest perfect square here, giving 2√2.
• Part (ii): the differences are 4 and -4, and the signs vanish on squaring, so the length is √32=4√2.
• Part (iii): treat the letters like numbers. The point (-a,-b) is the reflection of (a,b) through the origin, so the segment is twice the distance of (a,b) from the origin.
• Why surds: a neat surd answer tells the examiner you handled the radical correctly, which earns more than a rounded decimal in a proof-style question.

2√2, 4√2 and 2√a²+b² units respectively.

Spot the Pythagorean triple. Trained eyes read (36,15) as a scaled standard triple, so the distance comes out at sight.

• Factor: pull the common 3 out, since 36=3× 12 and 15=3× 5, leaving the clean leg pair (12,5).
• Recognise: the pair (12,5) has hypotenuse √12²+5²=13, one of the standard right-triangle triples.
• Scale back: multiply by the factor 3 to get the distance 3× 13=39 km, with √1521 used only as a check.
• Why it helps: knowing the common triples saves time in the board exam, where many distance questions are built from them on purpose, and the straight 39 km beats the 51 km road path.

39 units, so towns A and B are 39 km apart.

A faster slope check. Computing three surds is slow, so compare the slopes of two segments instead with only subtraction and division.

• Slope of AB: the rise over run from A to B is 3-52-1=-2, a clean whole number.
• Slope of AC: the rise over run from A to C is -11-5-2-1=163≈ 5.33, a different value.
• Verdict: the slopes differ, so the two segments point in different directions from the shared point A. The three points are therefore not collinear, agreeing with the distance test.
• The rule: equal slopes through a shared point mean collinear, while unequal slopes mean a genuine triangle. Both methods score, but the slope check is quicker when the numbers are ugly under the root.

Not collinear, confirmed by unequal slopes -2 and 163.

Symmetry tells the story. The vertex B sits exactly above the mid-point of the base AC, which proves the triangle isosceles at sight.

• Find the base mid-point: the mid-point of A(5,-2) and C(7,-2) is (5+72,-2-22)=(6,-2).
• Spot the alignment: the vertex B(6,4) lies directly above (6,-2) on the vertical line x=6, which is the perpendicular bisector of the base AC.
• Conclude: any point on the perpendicular bisector of AC is equidistant from A and C, so the two slant sides are equal without any further calculation and the triangle is isosceles.
• Carry it forward: this perpendicular-bisector idea reappears whenever a question asks for a point equidistant from two others, so it is worth recognising early.

Isosceles: B lies on the perpendicular bisector of AC, so AB=BC=√37.

Why both checks are needed. Equal sides alone do not prove a square, because a tilted rhombus also has four equal sides.

• Equal sides: all four sides came out as 3√2, which only guarantees a rhombus at this stage and nothing more.
• Diagonals decide: in a rhombus the diagonals are usually unequal, and they become equal only when every angle is a right angle, that is, when the figure is actually a square.
• Apply it here: the diagonals come out equal, so they force right angles and the figure is a square, which means Chameli is mistaken and Champa is right.
• The trap: a common board mistake is to give four equal sides and stop, calling the figure a square too soon. Always finish with the diagonal comparison to separate rhombus from square.

Square confirmed by equal sides and equal diagonals; Champa is right.

Always run the collinearity check first. Part (ii) is the trap of this question, and skipping the check is how most marks are lost here.

• Part (i): the four equal sides together with the equal diagonals give a square in just two short comparison steps, since nothing else has both properties at once.
• Part (ii): the sum of two short distances equals the third, which exposes that one vertex sits on the segment joining the other two. With three points falling in a straight line there is no enclosed region, so there is no quadrilateral to name at all.
• Part (iii): matching opposite sides but unequal diagonals is the exact signature of a parallelogram, and the unequal diagonals are what separate it from a rectangle, which would need equal diagonals instead.
• Where the marks are: the reasoning marks come from the comparison sentence and not from the arithmetic. State plainly which sides or diagonals match, or fail to match, before you name each figure, and never compute four sides without checking collinearity first.

(i) square, (ii) no quadrilateral, (iii) parallelogram, each justified by side and diagonal comparison.

Check the answer by back-substitution. Marks slip away when a sign is dropped, so verify the point really is equidistant before moving on.

• Distance to A: substituting gives √(-7-2)²+(0+5)²=√106 for the first point.
• Distance to B: substituting gives √(-7+2)²+(0-9)²=√106 for the second point.
• Match: both lengths come out equal, so the point is confirmed equidistant from the two given points as required.
• The geometry: the answer is where the perpendicular bisector of the segment crosses the x-axis. A quick back-check like this guards against arithmetic slips under exam pressure.

(-7,0), verified since both distances equal √106.

Read it as a right triangle. The horizontal gap is fixed and the hypotenuse is given, so the vertical gap is the one missing leg.

• Fixed sides: the horizontal leg is 10-2=8 and the hypotenuse is the given distance 10, both known up front.
• Missing leg: by Pythagoras the vertical leg is √10²-8²=√36=6, the only unknown side.
• Two answers: starting from y=-3 you can move 6 units either way, giving y=3 going up or y=-9 going down.
• Why two: this is the scaled three-four-five triple, which makes the vertical gap of 6 obvious and the two-answer structure natural, since the point can sit above or below the start.

y=3 or y=-9.

Separate the parts the question asks for. There are three deliverables here, so treat the two signs of x as separate scenarios.

• The values of x: equidistance gives x²=16, so x=4 or x=-4. Both are valid because the problem does not restrict the sign of the unknown.
• The length QR: this stays √41 for either sign, because only the square of x enters the expression and the sign is lost on squaring.
• The length PR: this uses x directly, so it differs between the two cases. Listing a value for each sign keeps the case work complete and avoids dropping marks.
• The general lesson: whenever a squared variable yields plus-or-minus values, scan the remaining quantities to see which depend on x and which depend only on its square. That tells you at once which answers split into cases.

x=± 4, QR=√41, and PR=√82 or 9√2 for x=4 or x=-4.

Cross-check using the midpoint. The answer line must pass through the mid-point of the segment, which gives a fast verification.

• Find the mid-point: the mid-point of the two given points is (3-32,6+42)=(0,5).
• Test the relation: substituting into 3x+y=5 gives 3(0)+5=5, which holds, so the bisector passes through the mid-point exactly as it must.
• Check the slopes: the slope of the segment is 4-6-3-3=13, while the answer line has slope -3. Their product is -1, which confirms the two are perpendicular.
• Why it proves it: the mid-point lying on the answer and the slopes multiplying to give the perpendicular condition together prove the relation is the correct perpendicular bisector, which is exactly what equidistant demands.

3x+y=5, the perpendicular bisector of the segment joining (3,6) and (-3,4).

Sanity-check the position. A two-to-three split puts the point a little nearer the first end, so its coordinates should lean that way.

• Read the ratio: the two parts total 5, so the point sits two-fifths of the way along from the first end toward the second.
• Walk the fraction: moving two-fifths along gives x=-1+25(4-(-1))=1 and y=7+25(-3-7)=3.
• Confirm: this fraction-of-the-way method lands on the same point as the section formula, so the two agree.
• Why bother: reading the ratio as a fraction of the journey is a quick mental check that the point lands on the correct side of the segment and not beyond either end.

(1,3), lying two-fifths of the way from (-1,7) to (4,-3).

Picture the two cut points. The first cut sits one-third along the segment and the second sits two-thirds along, so both coordinates should step evenly between the ends.

• Step the x: the full change in x is -6, so each third changes it by -2. Starting at 4 the cuts fall at 2 and then 0, ending at -2, which matches the computed values.
• Step the y: the full change in y is -2, so each third changes it by -23. Starting at -1 the cuts fall at -53 and then -73.
• Pair them: reading the two coordinate lists together gives the same two trisection points found by the section formula.
• The hallmark: equal steps in both coordinates are the signature of trisection, and checking them catches any ratio mix-up at once before it costs marks.

(2,-53) and (0,-73).

Set up the model carefully. The only real work is turning the story into points; after that the formulas are mechanical.

• Green flag: one-quarter of the 100 pots is 25 m, paired with line 2, which fixes the green flag at (2,25).
• Red flag: one-fifth of the same length is 20 m, paired with line 8, which fixes the red flag at (8,20).
• Apply the formulas: the horizontal gap is 6 and the vertical gap is 5, so the flags are √61 m apart, and the mid-point places the blue flag squarely between the two.
• Read the answer: a halfway flag always lands at the mid-point, so it sits on line 5 at the listed height, equally far from both runners. Careful modelling of the words earns the method marks here, not the arithmetic.

Flags are √61 m apart; the blue flag is at (5, 22.5).

Confirm with the y-coordinate. Finding the ratio from x alone scores, but the y-coordinate proves the point lies on the segment.

• Set up: with the ratio fixed, the y-coordinate from the section formula is k(-8)+10k+1=27(-8)+1027+1.
• Simplify: the top reduces to 547 and the bottom to 97 after clearing the small fractions.
• Divide: the two parts cancel to give 6, which matches the given y-coordinate, so the ratio is correct.
• The habit: checking the second coordinate is the standard guard against a point that satisfies x but lies off the line. Always run it when the ratio is the final answer.

Ratio 2:7, confirmed by the matching y-coordinate 6.

Read the sign change as a clue. One end sits below the axis and the other an equal amount above it, so the crossing must be halfway.

• Spot the symmetry: one point is 5 below the x-axis and the other is 5 above it, so the axis sits midway between them in the vertical direction.
• Predict the ratio: midway means a ratio of one-to-one, which the equation 5k-5=0 confirms with the value k=1.
• Find the point: the mid-point x-coordinate is 1+(-4)2=-32, giving the crossing point on the axis.
• The shortcut: whenever the two y-values are equal and opposite, the axis always bisects the segment, so you can predict the one-to-one ratio and use the calculation purely as a check.

Ratio 1:1; point of division (-32,0).

Why the diagonal property wins here. The mid-point method needs only one equation per unknown and avoids the side-vector notation that this class has not yet covered.

• Shared centre: the two diagonals meet at one common mid-point, found from the known coordinates of three vertices, so both diagonals must pass through it.
• Solve for x: forcing the first diagonal to pass through that shared centre gives a single equation that fixes the value of x.
• Solve for y: forcing the second diagonal through the same centre gives a second equation that fixes the value of y.
• The takeaway: the single fact that diagonals bisect each other resolves both unknowns cleanly, which is why it is the preferred approach for parallelogram problems at this level.

x=6, y=3, from the equal mid-points of the diagonals.

Visualise the centre as the balance point. The centre is equally far from both ends of the diameter, so one step repeated lands on the far end.

• First step: go from the known end to the centre, a change of one across and seven down between the two points.
• Repeat it: apply the same change again starting from the centre, which lands exactly on the unknown far end.
• Cross-check: that endpoint agrees with the value found by the mid-point calculation, so both methods confirm each other.
• Why it works: seeing the centre as the mid-point, and a diameter as two equal steps, turns this into a one-move problem that doubles as a built-in check.

A(3,-10).

Use the fraction-of-the-way form. Because the point sits a fixed fraction of the way along, you can add that fraction of the coordinate change directly to the start.

• The change: the move from start to end is (2-(-2), -4-(-2))=(4,-2), the total shift across the segment.
• Scale it: take three-sevenths of that shift to get (127,-67), the partial move to the point.
• Add to the start: adding this to the starting point lands exactly on the required point.
• Why use it: this method skips converting the fraction to a ratio altogether, yet lands on the same point, which makes it a reliable double-check on the section-formula answer.

P=(-27,-207).

Equal steps confirm the answer. Four equal parts mean each coordinate rises by the same amount at every cut from start to end.

• Step the x: the total x-change of 4 over four parts makes each step +1, so the values run -2,-1,0,1,2 and the three interior values are -1,0,1.
• Step the y: the total y-change of 6 over four parts makes each step +32, so the interior values are 72,5,132.
• Pair them: matching the interior x and y values in order gives the three division points, the same ones the mid-point method produces.
• Why it is fast: reading the points as a steady progression in both coordinates lists all three at once and verifies them in a single glance.

P(-1,72), Q(0,5), R(1,132).

Use the diagonals, not the four sides. The hint points straight to half the product of the diagonals, so find those two lengths and skip the longer route through equal sides and an angle.

• First diagonal: the length 4√2 links the opposite vertices, jumping across the figure rather than along an edge.
• Second diagonal: the length 6√2 links the other pair of opposite vertices in the same way.
• Combine: half the product of the two lengths gives the area as 24 square units, the final answer.
• The one decision: the only judgement call is picking the diagonals as opposite vertices rather than adjacent ones. Once chosen, the surds collapse neatly since root two times root two is two.

24 square units.